If the equation of the plane passing through the point (1,1,2) and perpendicular to the line x-3 y+ 2 z-1=0=4 x-y+z is A t+ B y+ C z=1, then 140( C - B + A ) is equal to

Question

If the equation of the plane passing through the point (1,1,2) and perpendicular to the line x-3 y+ 2 z-1=0=4 x-y+z is A t+ B y+ C z=1, then 140( C - B + A ) is equal to (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

x-3 y+2 z-1=0 4 x-y+z=0 n 1 × n 2=| hat i hat j hat k 1 -3 2 4 -1 1| =- hat i +7 hat j +11 hat k Dr of normal to the plane is -1,7,11 Equation of plane: -1(x-1)+7(y-1)+11(z-2)=0 -x+7 y+11 z=28 (-1/28) x+(7 y/28)+(11 z/28)=1 Ax + By + Cz =1 140( C - B + A )=140((11/28)-(7/28)-(1/28)) =140 × (3/28)=15

Q. If the equation of the plane passing through the point (1,1,2) and perpendicular to the line x−3y+ 2z−1=0=4x−y+z is At+By+Cz=1, then 140(C−B+A) is equal to_______

Q. If the equation of the plane passing through the point $(1, 1, 2)$ and perpendicular to the line $x - 3 y +$ $2 z - 1 = 0 = 4 x - y + z$ is $A t + B y + C z = 1$ , then $140 (C - B + A)$ is equal to_______